$ C = \left[\begin{array}{rrr}5 & 0 & -2 \\ 3 & 0 & 1\end{array}\right]$ $ F = \left[\begin{array}{rr}1 & 2 \\ 4 & -2 \\ -1 & -1\end{array}\right]$ What is $ C F$ ?
Solution: Because $ C$ has dimensions $(2\times3)$ and $ F$ has dimensions $(3\times2)$ , the answer matrix will have dimensions $(2\times2)$ $ C F = \left[\begin{array}{rrr}{5} & {0} & {-2} \\ {3} & {0} & {1}\end{array}\right] \left[\begin{array}{rr}{1} & \color{#DF0030}{2} \\ {4} & \color{#DF0030}{-2} \\ {-1} & \color{#DF0030}{-1}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ C$ , with the corresponding elements in column $j$ of the second matrix, $ F$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ C$ with the first element in ${\text{column }1}$ of $ F$ , then multiply the second element in ${\text{row }1}$ of $ C$ with the second element in ${\text{column }1}$ of $ F$ , and so on. Add the products together. $ \left[\begin{array}{rr}{5}\cdot{1}+{0}\cdot{4}+{-2}\cdot{-1} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ C$ with the corresponding elements in ${\text{column }1}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{5}\cdot{1}+{0}\cdot{4}+{-2}\cdot{-1} & ? \\ {3}\cdot{1}+{0}\cdot{4}+{1}\cdot{-1} & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ C$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ F$ and add the products together. $ \left[\begin{array}{rr}{5}\cdot{1}+{0}\cdot{4}+{-2}\cdot{-1} & {5}\cdot\color{#DF0030}{2}+{0}\cdot\color{#DF0030}{-2}+{-2}\cdot\color{#DF0030}{-1} \\ {3}\cdot{1}+{0}\cdot{4}+{1}\cdot{-1} & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{5}\cdot{1}+{0}\cdot{4}+{-2}\cdot{-1} & {5}\cdot\color{#DF0030}{2}+{0}\cdot\color{#DF0030}{-2}+{-2}\cdot\color{#DF0030}{-1} \\ {3}\cdot{1}+{0}\cdot{4}+{1}\cdot{-1} & {3}\cdot\color{#DF0030}{2}+{0}\cdot\color{#DF0030}{-2}+{1}\cdot\color{#DF0030}{-1}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}7 & 12 \\ 2 & 5\end{array}\right] $